Question

Mary told John her score on the American High School Mathematics Examination (AHSME), which was over $80.$ From this, John was able to determine the number of problems Mary solved correctly. If Mary’s score had been any lower, but still over $80,$ John could not have determined this. What was Mary’s score? (Recall that the AHSME consists of $30$ multiple-choice problems and that one’s score, $s,$ is computed by the formula $s=30+4c-w,$ where $c$ is the number correct and $w$ is the number wrong; students are not penalized for problems left unanswered.)

Answer

Answer: 119

Concepts:Diophantine Equation counting integers in a range

Difficulty rating: 2360

Small Hint:

For a fixed score $s,$ express the number wrong in terms of $s$ and $c$

Big Hint:

Use $w\geq0$ and $c+w\leq30$ to bound the possible integer values of $c$

Solution:

From $s=30+4c-w,$ we have $w=30+4c-s.$ The conditions $w\geq0$ and $30-c-w\geq0$ give $\left\lceil\frac{s-30}{4}\right\rceil \leq c\leq \left\lfloor\frac{s}{5}\right\rfloor.$ Evaluating these integer endpoints for scores $81$ through $118$ always leaves at least two possible values of $c.$ At $s=119,$ both endpoints equal $23,$ so John can determine $c=23.$ Thus the first score over $80$ with the required property is $119.$

1984 AIME Problem 10

Answer: 119

Difficulty rating: 2360

Small Hint:

Big Hint:

Solution:

Problem 10 in Other Years